Question

How do you differentiate `y=(x+1)^2(2x-1)`?

Answer 1

`dy/dx =4x(2x+1)`

Explanation:

`y=(x+1)^2(2x+1)`

Use the product rule and chain rule to differentiate `y`

Product rule:

`d/dx(pq) = qp' + pq'`

Chain rule:

`d/dx([f(x)]^n) = n[f(x)]^(n-1)f'(x)`

`dy/dx = 2(x+1)(2x-1) + 2(x+1)= 2(x+1)(2x-1+1) = 4x(2x+1)`

Answer 2

`6x(x + 1)`

Explanation:

Actually, there are mainly 2 methods of solving your question,

1. Either simplifying the equation and differentiating it

2. Or differentiating first and then simplifying it.

Obviously, the first one is the easiest for beginners,

`rArr y = (x + 1)^2(2x - 1)`

Simplifying this, I get,
`rArr y = 2x^3 + 3x^2 - 1`

Now,
`rArr dy/dx = 6x^2 + 6x`

`therefore, dy/dx = 6x(x + 1)`

ENJOY MATHS !!!!!!!!!